A Spherical balloon is being inflated at the rate of 35cc/min. The rate of increase in the surface area(in cm2/min.) of the balloon when its diameter is 14 cm, is

Options

A. 10
B. `sqrt(10)`
C. 100
D. `10 sqrt(10)`

Answer

A

Explanation

Let volume of spherical ballon `V=4/3pir^(3)`
`implies(dV)/(dt)=4/3 pi 3r^(2) (dr)/(dt)implies35=4pi(7)^(2)(dr)/(dt)`
[ `:'` Given `r=7cm` and `(dv)/(dt)=35cm^(3)min^(-1)`]
`implies(dr)/(dt)=5/(28pi)`
Surface area of spherical balloon `S=4pir^(2)`
`:.(dS)/(dt)=8pir(dr)/(dt)=8pixx7xx5/(28pi)=10cm^(2)min^(-1)`